The real (that is, the
non-complex)
zeroes of a polynomial correspond to the x-intercepts
of the graph of that polynomial. So you can find information about the
number of real zeroes of a polynomial by looking at the graph, and conversely
you can tell how many times the graph is going to touch or cross the x-axis
by looking at the zeroes of the polynomial (or at the factored form of
the polynomial).

A zero has a "multiplicity",
which refers to the number of times that its associated factor appears
in the polynomial. For instance, the quadratic (x
+ 3)(x – 2) has
the zeroes x
= –3 and x
= 2, each occuring once.
The eleventh-degree polynomial
(x + 3)4(x – 2)7
has the same zeroes as did the quadratic, but in this case, the x
= –3 solution has multiplicity
4
because the factor (x
+ 3) occurs four times
(the factor is raised to the fourth power) and the x
= 2 solution has multiplicity
7
because the factor (x
– 2) occurs seven times.

All four graphs have the
same zeroes, at x
= –6 and at x
= 7, but the multiplicity
of the zero determines whether the graph crosses the x-axis
at that zero or if it instead turns back the way it came.

The following graph
shows an eighth-degree polynomial. List the polynomial's zeroes with
their multiplicities.

I can see from the graph
that there are zeroes at x
= –15, x = –10, x = –5, x = 0, x = 10,
and x
= 15, because the graph
touches or crosses the x-axis
at these points. (At least, I'm assuming that the graph crosses at exactly
these points, since the exercise doesn't tell me the exact values. When
I'm guessing from a picture, I do have to make certain assumptions.)

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Since the graph just touches
at x
= –10 and x
= 10, then it must
be that these zeroes occur an even number of times. The other zeroes
must occur an odd number of times. The odd-multiplicity zeroes might
occur only once, or might occur three, five, or more times each; there
is no way to tell from the graph. (At least, there's no way to tell
yet — we'll learn more about that on the next page.) And the
even-multiplicity zeroes might occur four, six, or more times each;
I can't tell by looking.

But if I add up the minimum
multiplicity of each, I should end up with the degree, because otherwise
this problem is asking for more information than is available for me
to give. I've got the four odd-multiplicity zeroes (at x
= –15, x = –5, x = 0, and
x
= 15) and the two even-multiplicity
zeroes (at x
= –10 and x
= 10). Adding up their
minimum multiplicities, I get 1
+ 2 + 1 + 1 + 2 + 1 = 8,
which is the degree of the polynomial. So the minimum multiplicities
are the correct multiplicities.

I was able to compute the
multiplicities of the zeroes in part from the fact that the multiplicities
will add up to the degree of the polynomial, or two less, or four less,
etc, depending on how many complex
zeroes there might be. But multiplicity problems don't usually get into
complex numbers.